I'm trying to prove that for a transcendental number $a$ the module $\mathbb{Z}[a]$ over $\mathbb{Z}[(a+1)^2]$ is free. For $\mathbb{Z}[a+1]$ over $\mathbb{Z}[(a+1)^2]$, the basis is $\{1,a+1\}$. What I wanted to show now, is that because every polynomial from $\mathbb{Z}[a]$ can be written as a polynomial from $\mathbb{Z}[a+1]$ (this requires solving a system of linear equations for the coefficients), it has the same basis. Is this argument correct?